A s-block element (M) reacts with oxygen to f | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The reaction of alkali metals with excess oxygen leads to the formation of superoxides $(MO_{2})$.Potassium $(K)$,Rubidium $(Rb)$,and Cesium $(Cs)

Vedclass · Accepted Answer

$K$

Vedclass · Answer

(B) The reaction of alkali metals with excess oxygen leads to the formation of superoxides $(MO_{2})$.Potassium $(K)$,Rubidium $(Rb)$,and Cesium $(Cs)$ form superoxides ($KO_{2}$,$RbO_{2}$,$CsO_{2}$) which are paramagnetic due to the presence of one unpaired electron in the $\pi^* 2p$ molecular orbital of the superoxide ion $(O_{2}^{-})$.$KO_{2}$ is a pale yellow solid.Sodium $(Na)$ forms a peroxide $(Na_{2}O_{2})$,which is diamagnetic and white.Calcium $(Ca)$ and Magnesium $(Mg)$ form oxides ($CaO$,$MgO$) or peroxides $(CaO_{2})$,which are diamagnetic.Therefore,the element $(M)$ is $K$.

$A$ $s$-block element $(M)$ reacts with oxygen to form an oxide of the formula $MO_{2}$. The oxide is pale yellow in colour and paramagnetic. The element $(M)$ is:

Similar Questions

The ease of adsorption of the hydrated alkali metal ions on an ion-exchange resins follows the order:

For alkali metals, which one of the following trends is incorrect?

What are the metal ions present in carnallite?

The basicity of the hydroxides of the following alkali metals is of the order:

Explain why is $sodium$ less reactive than $potassium$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module